Properties of Gases
CHAPTER 11
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
1
CHAPTER 11 Learning Objectives
 Describe properties of a gas
 Read a barometer & monometer
 Unit conversions for moles, temperature, pressure, and volume
 Explain relationships between variables of state & predict effect of a
change to a system
 Apply gas law to stoichiometry, molecular weight, and density
problems
 Understand the relationship between variables of state in terms of
Kinetic Molecular Theory
 Calculate mole fractions and partial pressures
 Compare rates of effusion
 Compute variables of state using the real gas law
2
CHAPTER 11 Lecture Road Map
① What is a gas?
②Define & measure variables of state
③Relationships between variables of state
④Equation of state for an ideal gas
⑤Dalton’s law of partial pressure & gas stoichiometry
⑥Kinetic Molecular Theory & Graham’s law
⑦Real gas law
3
Avagadro’s
Law
Mole Fraction
& Mole %
Guy-Lussacs
Law
Charles Law
Boyles Law
Ideal Gas
Law
Volume
Dalton’s Law
Pressure:
Barometers & Monometers
CHAPTER 11
Properties of Gases
Real Gas
Law
Pressure
Variables of
State
Temperature:
K, °C, °F
Kinetic
Molecular
Theory
Relationships
between
variables of state
Graham’s Law
Absolute 0
4
Group
Problem
In groups of 3-5 brainstorm
how to describe a gas.
What are some observable properties?
What variables would you use to describe a gas?
5
Group
Problem
Describe a gas:
o Will expand to fill a volume
o Mostly empty space so can be compressed
o Can expand & contract with temperature
o Particles constantly in motion & constantly colliding
o Some gases are heavier then others and sink to the
floor rather then rise to the ceiling
6
Variable
s of
State
Physical Properties of Gases
Despite wide differences in chemical properties, all gases
more or less obey the same set of physical properties
1. Pressure (P )
2. Volume (V )
3. Temperature (T )
4. Amount = moles (n)
7
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Volume (V)
V=l×w×h
for a cube
V = (4/3) π r3
for a sphere
V = π r2 h
for a cylinder
Units of Volume: Liters (L)
1 L = 0.001 m3
= 1000 cm3
= 1000 mL
8
Variable
s of
State
# of Moles (n)
Avagadro’s number (NA) allows us to measure the number of
particles of a gas as the number of moles:
NA = 6.02214129 × 1023 particles/mole
We can measure the number of moles of a gas by measuring its
mass and knowing its Molar Mass
Molar Mass = mass / (# of moles)
M = m/n
9
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
How many moles of the CFC pollutant CCl2F2 are in 50.0g?
10
Group
Problem
Calculate the mass of 3 moles of nerve agent VX:
CH3CH2
11
Variable
s of
State
Temperature (T)
Temperature is measured with a thermometer
usually in °C, °F, or Kelvin.
0°C = 273 K
1°C = (1°F -32) × (5/9)
12
Group
Problem
If Room Temperature (RT) is 25°C,
what is RT in Kelvin? °F?
Write out a formula to convert °F to K.
13
Group
Problem
If Room Temperature (RT) is 25°C,
what is RT in Kelvin? °F?
RT = 25°C + 273 K = 298 K
14
Variable
s of
State
Pressure (P)
force
Pressure =
area
Force = mass × area
o Pressure is the force of the collisions of the gas
distributed over the surface area of the container walls
o Earth exerts gravitational force on everything with
mass near it
o Atmospheric Pressure of earth: gravity pulling on
gases creating a blanket around earth
15
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
Calculate Atmospheric Pressure on Earth.
① Identify information needed
② Research
③ Solve
16
Group
Problem
Calculate Atmospheric Pressure on Earth.
17
Variable
s of
State
Pressure (P)
A vacuum exerts zero pressure on a containers walls.
18
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Pressure (P)
Measure Atmospheric
pressure with a barometer.
Toricelli Barometer:
o Tube that is 80 cm in
length
o Sealed at one end
o Filled with mercury
o In dish filled with mercury
19
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Pressure (P)
o Atmospheric pressure
o Pushes down on mercury
o Forces mercury up tube
o Weight of mercury in tube
o Pushes down on mercury
in dish
o When two forces balance
o Mercury level stabilizes
o Read atmospheric
pressure
20
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Pressure (P)
o If atmospheric pressure is high
o Pushes down on mercury in
dish & increase level in tube
o If atmospheric pressure is low
o Pressure on mercury in dish
less than pressure from
column & decrease level in
tube
Therefore:
o Height of mercury in tube is the
atmospheric pressure
21
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Pressure (P)
measured with a barometer
P=g×d×h
d=density of the liquid
g= gravitational
acceleration
h=height of the column
supported
22
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Pressure (P)
Typical range of pressure for most places where people live
730 to 760 mm Hg
Top of Mt. Everest
Atmospheric Pressure = 250 mm Hg
Standard Atmosphere (atm)
Average pressure at sea level
Pressure needed to support column of mercury 760 mm high
measured at 0 °C = 1 atm
23
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Pressure (P)
SI unit for pressure
Pascal = Pa = 1 N/m2
1atm = 101,325 Pa = 101 kPa
100 kPa = 0.9868 atm
Other units of pressure
1.013 Bar = 1013 mBar = 1 atm
760 mm Hg = 1 atm
760 torr = 1 atm
At sea level 1 torr = 1 mm Hg
24
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
Express Pressure in atm and kPa for a gas at 705 mmHg.
25
Variable
s of
State
Pgas = Patm
Pressure (P)
Open Ended Manometer
Pgas > Patm
Pgas < Patm
Gas pushes mercury
up tube
Atmosphere pushes
mercury down tube
26
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
CO2 collected in a monometer in a
lab with a barometric reading of 97
kPa. What is the Pressure of CO2?
33 mm
27
Variable
s of
State
Pressure (P)
Closed-end Manometer
o Arm farthest from
vessel (gas) sealed
o Tube filled with
mercury
o Then open system to
flask and some
mercury drains out of
sealed arm
o Vacuum exists above
mercury in sealed arm
28
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Variable
s of
State
Pressure (P)
Closed-end Manometer
o Level of mercury in arm
falls, as not enough
pressure in the flask to hold
up Hg
o Patm = 0
o Pgas = PHg
o So directly read pressure
29
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
What is the pressure of an unknown gas
within this closed monometer?
Closed monometer
437 mm
205 mm
30
Ideal
Gas Law
Boyle’s Law
Volume will change to equalize pressure with
atmosphere is not in a rigid vessel.
V α 1/P
31
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Ideal
Gas Law
Charles Law
If Pressure is constant but freeze a balloon, it decreases in32V
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Ideal
Gas Law
Charles Law
If Pressure is constant but freeze a balloon, it decreases in V
VαT
33
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Ideal
Gas Law
Gay Lussac’s Law
P µT
o Volume (V ) and number of moles (n) are constant
o P increases as T increases
Low T, Low P
o Showed that gas pressure
is directly proportional
to absolute temperature
P
High T, High P
T (K)
34
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
Force of Collisions
P
Area
What happens to gas pressure when you raise the
temperature?
If the container can expand in
response to the force
No change in pressure is
observed because the
area increased.
In a rigid walled container
Pressure increases
because the faster
moving molecules hit
the walls of the
container with greater
force
35
Ideal
Gas Law
Combined Gas Law
1
o Boyle’s law: P 
V
o Charles Law:
T V
o Guy-Lussac’s Law:
o
T
P
V
T P
is equivalent to
o For any two conditions:
T
P
V
PV
=C
T
P1V1 P2V2

T1
T2
36
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Ideal
Gas Law
Combined Gas Law
P1V1
T1
=
Boyle’s Law
T1 = T2
Charles’ Law
P1 = P2
Gay-Lussac’s V1 = V2
Law
P2V2
T2
P1V1 = P2V2
V1
T1
P1
T1
=
=
V2
T2
P2
T2
37
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
N2 + H2  NH3
How much H2 at 0°C and 0.86 atm do you need to react
completely with 750 mL of N2 at 1.5 atm and 20°C to
form ammonia?
What is the number of moles of ammonium produced if
the density of hydrogen is 0.08988 g/L?
Hint: is this equation balanced?
38
Group
Problem
A sample of helium gas occupies 500.0 mL at 1.21 atm
Calculate the volume of the gas if the pressure is
reduced to 491 torr
Ideal
Gas Law
Avagadro’s Law
Vαn
At standard temperature (273 K)
And standard pressure (1 atm)
1 mole of any gas will occupy the same volume
40
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
Force of Collisions
P
Area
What happens to gas pressure when you increase the
number of molecules in the container?
If a container can
expand
No pressure
change is
observed.
In a rigid walled
container
pressure increases
because more
molecules hit the
walls of the container,
thus exert a greater
force on the container
41
Ideal
Gas Law
Putting It All Together
If
PV
na
T
then
PV
R=
Tn
o R is the universal gas constant
42
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
Plug in values of T, V, n and P for 1 mole of gas at STP (1
atm and 0.0 °C)
T = 0.0 °C = 273.15 K
P = 1 atm
V = 22.4 L
n = 1 mol
PV
1 atm ´ 22.4 L
R=
=
nT 1 mol ´ 273.15 K
R = 0.082057 L atm mol–1 K–1
43
Ideal
Gas Law
Putting It All Together
If
PV
na
T
then
PV
R=
Tn
o R is the universal gas constant
o R = 0.0821 (L×atm) / (mol×K)
= 8.314 J / (mol×K)
= 8.314 (kg×m2) / (s2×mol×K)
PV = nRT
44
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
Calculate Molar Volume = the volume 1 mole of
any gas occupies at 1 atm and 273 K
45
Group
Problem
Compare Molar volume at STP to Room
Temperature(25°C) assuming pressure remains
constant
46
Group
Problem
How many liters of N2(g) at 1.00 atm and 25.0 °C
are produced by the decomposition of 150. g of
NaN3?
2NaN3(s)  2Na(s) + 3N2(g)
47
Group
Problem
At what temperature will 1.50 moles of CH4 occupy a
1 L container at 10atm?
48
Group
Problem
o PV = nRT
od=m/V
oM=m/n
Write out the ideal gas law in terms of
density & then in terms of molar mass
49
Ideal
Gas Law
Considering Density & Molar Mass
o PV = nRT
od=m/V
oM=m/n
P (m / d) = nRT
PV = (m / M) RT
RT / P = M / d
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
50
Group
Problem
At what temperature will 479.4 g of Br2 occupy a 1 L
container at 25 atm?
51
Group
Problem
2H2O (l) + catalyst + hv  2H2 (g) + O2 (g)
Under the following conditions if 40.53 g of a 1 L
Volume of water is split into hydrogen and oxygen gas,
what is the volume the gas mixture if collected in a
balloon (ignore water vapor)?
T = 25°C
Patm = 1.025 atm
dH2O = 1 g/mL
mH2O = 40.53 g
52
Dalton’s
Law
Partial Pressure
o For mixture of non-reacting gases in container
o Total pressure exerted is sum of the individual partial pressures
that each gas would exert alone
o Ptotal = Pa + Pb + Pc + ···
o Where Pa, Pb, and Pc are the partial pressures
o Partial pressure
o Pressure that particular gas would exert if it were alone in
container
53
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Dalton’s
Law
Partial Pressure
Assuming each gas behaves ideally partial pressure of each
gas can be calculated from ideal gas law
Pa =
naRT
Pb =
V
nbRT
Pc =
V
ncRT
V
So total pressure is
Ptotal = Pa + Pb + Pc + × × ×
=
naRT
V
+
nbRT
V
+
ncRT
V
+×××
54
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Dalton’s
Law
Partial Pressure
Rearranging
Ptotal
æ RT
= na + nb + nc + × × × çç
èV
Or
Ptotal
æ RT
= ntotal çç
èV
(
)
ö
÷÷
ø
ö
÷÷
ø
Where ntotal = na + nb + nc + ···
ntotal = sum of number moles of various gases in mixture
55
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
2H2O (l) + catalyst + hv  2H2 (g) + O2 (g)
With an excess of water (1L) we know that 40.53 g (or
2.25 moles) of water will split. what is the partial
pressure of H2 gas and the partial pressure of O2 gas if
it is collected in a 25L rigid container (ignore water
vapor & change in water volume).
T = 25°C
Patm = 1.025 atm
dH2O = 1 g/mL
VH2Ototal = 1L
Vcontainer = 25L
nH2O = 2.25 moles
56
Dalton’s
Law
Vapor Pressure
Collected gas pressure must be corrected for water
vapor
Ptotal=Pgas + Vpwater
57
Group
Problem
32.5 mL of Hydrogen gas is collected over water at 25
ºC and 755 torr. What is the pressure of dry
hydrogen gas? (VP25ºC = 23.76 mmHg)
Correct Pt to find the Pdry gas:
755-23.76 torr=731.24 torr
731 torr = Phydrogen
58
Group
Problem
2H2O (l) + catalyst + hv  2H2 (g) + O2 (g)
Continuing the previous problem, what is the total
pressure of the system if we include water vapor
pressure (still ignoring change in volume of water)?
T = 25°C
Patm = 1.025 atm
dH2O = 1 g/mL
VH2Ototal = 1L
Vcontainer = 1.25L
mH2O = 40.53 g (that reacts)
59
Dalton’s
Law
Mole Fraction
Mole Fraction (χ)
Ratio of number moles of given component in mixture to total
number moles in mixture
cA =
nA
n A + nB + nC + × × × + nZ
=
nA
n total
Mole % = c A ´ 100%
60
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
2H2O (l) + catalyst + hv  2H2 (g) + O2 (g)
Continuing the previous problem, what is mole fraction
of water vapor in the system?
T = 25°C
Patm = 1.025 atm
dH2O = 1 g/mL
VH2Ototal = 1L
Vcontainer = 1.25L
mH2O = 40.53 g (that reacts)
61
Dalton’s
Law
Mole Fraction
æV
n A = PA çç
è RT
• If V and T are constant then,
V
RT
ö
÷÷
ø
= constant
• For mixture of gases in one container
æV ö
PA çç
÷÷
è RT ø
XA =
æV ö
æV ö
æV
PA çç
÷÷ + PB çç
÷÷ + PC çç
è RT ø
è RT ø
è RT
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
ö
æV
÷÷ + × × × + PZ çç
ø
è RT
ö
÷÷
ø
62
Dalton’s
Law
V
RT
Mole Fraction
cancels, leaving
cA =
PA
PA + PB + PC + × × × + PZ
or
cA =
PA
Ptotal
=
nA
n total
63
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Dalton’s
Law
Mole Fraction
o Partial pressure of particular component of gaseous
mixture
o Equals mole fraction of that component times total
pressure
PA = c A ´ Ptotal
64
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
The total pressure of a 1 L container of a room
298K gas mixture is 628 torr. What is the
pressure of Cl2 if there are 20 mg of CO2 and 8
mg of Cl2?
65
Small Group Problems:
1. 22.4 L of He at 25 ºC are heated to 200.ºC. What is the resulting volume?
2. A sample of fluorine gas occupies 275 mL at 945 torr and 72 ºC. What is the
mass of the sample?
3. What is the density of NO2 at 200 ˚C and 600. torr?
4. What is the molar mass of a gas with a density of 6.7 g/L at -73.ºC and a pressure
of 2.49 atm?
5. A sample of oxygen gas occupies 500.0 mL at 722 torr and –25 ºC. Calculate the
temperature in ºC if the gas has a volume of 2.53 L at 491 mm Hg.
6. What is the molar mass of a sample of gas if 2.22 g occupies a volume of 5.0 L a
35 ºC and 769 mm Hg?
a.1.3 g/mol
b.0.015 g/mol
c.0.090 g/mol
d. None of these
7. What is the mole fraction of N2 in the atmosphere? 1.000atm Air = .7808 atm N2+
.2095 atm O2+ .0093 atm Ar + .00036 atm CO2
8. Pump 520 mm Hg N2 and 250 mm Hg O2 into an empty gas cylinder. What is the
overall pressure of the mixture?
66
Kinetic
Molecular
Theory
5 Assumptions
① Gas particles are tiny, their V is negligible.
② Particles travel in a straight line, in random
directions.
③ 0 intermolecular attraction.
④ Elastic collisions, no Energy is lost.
⑤ If KE α T, then assume average KE α T.
67
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
① Compare to observations from beginning of class.
② Describe the relationships of state variables in
terms of Kinetic Molecular Theory:
Pα1/V
PαT
VαT
Vαn
68
Kinetic
Molecular
Theory
Boyles Law
P = (nRT )
Boyle’s Law
1
V
o Decrease in V, means
gas particles hit wall more often
o Increase P
69
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Kinetic
Molecular
Theory
Guy-Lussac’s Law
Guy-Lussac’s Law
æ nR ö
P = çç ÷÷ T
èV ø
o As T increases
o KEave increase
o Speeds of molecules
increases
o Gas particles hit wall more
often as V same
o So P increase
70
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Kinetic
Molecular
Theory
Charles Law
Charles Law:
æ nR ö
V = çç ÷÷ T
èP ø
o As T increases
o KEave increases
o Speeds of molecules
increases
o Gas particles hit wall more
often as pressure remains
the same
o So volume increases
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
71
Kinetic
Molecular
Theory
Avagadro’s Law
o For ideal gas at constant T and P
o V is directly proportional to n
æ RT
V = çç
è P
ö
÷÷ n
ø
o Kinetic Theory of Gases account for this
o As the number of moles of gas particles increase at
same T
o Holding T and P constant
o Must V must increase
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
72
Kinetic
Molecular
Theory
Dalton’s Theory
Ptotal   Pindiv idual gases
o Expected from kinetic theory of gases
o All gas particles are independent of each other
o Volume of individual particles is unimportant
o Identities of gases do not matter
o Conversely, can think of Dalton's Law of Partial Pressures as
evidence for kinetic theory of gases
o Gas particles move in straight lines, neither attracting nor
repelling each other
o Particles act independently
o Only way for Dalton's Law to be valid
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
73
Kinetic
Molecular
Theory
Absolute 0
T  KE av e
1
 m( v 2 )
2
o If KEave = 0, then T must = 0.
o Only way for KEave = 0, is if v = 0 since
m  0.
o When gas molecules stop moving, then gas
as cold as it can get
o Absolute zero
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
74
Kinetic
Molecular
Theory
Derivation of PV = nRT &
Important equations:
o PV = nRT
o KE = (3/2) RT
75
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Kinetic
Molecular
Theory
Graham’s Law of Effusion
Effusion
Diffusion
Gas mixing through Vacuum
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Gases mixing
76
Kinetic
Molecular
Theory
Derivation of Graham’s Law of Effusion
We can describe how fast a gas will effuse
Some Important Equations:
o ūrms =√(3RT) / M
o (ūrms)A /(ūrms)B = √(MB /MA)
77
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Kinetic
Molecular
Theory
Graham’s Law of Effusion
1
Effusion Rate 
d
(constant P and T)
dB
Effusion Rate ( A )
dB


Effusion Rate (B )
dA
dA
And dA  MM
(constant V and n)
Effusion Rate (A)
dB
MB
=
=
Effusion Rate (B)
dA
MA
Therefore, heavier gases effuse slower then
lighter gases
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
78
Group
Problem
H2O
CO2
1m
Which gas will travel the farthest? How far will
the CO2 travel down the tube before meeting
the gaseous water?
79
Real
Gas Law
Review Assumptions
Kinetic Molecular Theory
& Ideal Gas Law
① Gas particles are tiny,
their V is negligible.
② Particles travel in a
straight line, in random
directions.
③ 0 intermolecular
attraction.
④ Elastic collisions, no
Energy is lost.
⑤ If KE α T, then assume
average KE α T.
Real Gas Law
Real gases do not obey the
ideal gas law!
80
Real
Gas Law
Experimental Data
① Gas molecules have
finite volumes
o They take up space
o Less space of kinetic
motions
o Vmotions < Vcontainer
o Particles hit walls of
container more often
o Pressure is higher
compared to ideal
81
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Real
Gas Law
Experimental Data
①D
② Particles do attract each other
o Even weak attractions means they hit walls of
container less often
o Therefore, pressure is less than ideal gas
82
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Real
Gas Law
Experimental Data
83
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Real
Gas Law
Review Assumptions
Kinetic Molecular Theory
& Ideal Gas Law
① Gas particles are tiny,
their V is negligible.
② Particles travel in a
straight line, in random
directions.
③ 0 intermolecular
attraction.
④ Elastic collisions, no
Energy is lost.
⑤ If KE α T, then assume
average KE α T.
Real Gas Law
① Gas particles do take up
Volume:
Vmeas – nb
② Attractive forces
between molecules exist
and effect a particle’s
path.
Pmeas + [(n2a) / V2]
84
Real
Gas Law
Van Der Waal’s Equation for Real Gases
æ
n 2a ö
ççP +
÷÷ V - nb = nRT
2
V ø
è
(
corrected P
)
corrected V
o a and b are van der Waal's constants
o Obtained by measuring P, V, and T for real gases
over wide range of conditions
o Table X.X. in your textbook.
85
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Group
Problem
Use the Real Gas Law to calculate the Vapor
Pressure of 1 mole water in a 1 Liter container at
the following Temperatures:
T
Pvapor
0°C
75°C
100°C
127°C
427°C
86
Avagadro’s
Law
Mole Fraction
& Mole %
Guy-Lussacs
Law
Charles Law
Boyles Law
Ideal Gas
Law
Volume
Dalton’s Law
Pressure:
Barometers & Monometers
CHAPTER 11
Properties of Gases
Real Gas
Law
Pressure
Variables of
State
Temperature:
K, °C, °F
Kinetic
Molecular
Theory
Relationships
between
variables of state
Graham’s Law
Absolute 0
87